What is the integral from 0 to 2 of (3x-1)^3 ?

The integral from 0 to 2 of (3x-1)^3 is 52

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integral from 0 to 2 of (3x-1)^3

\int_{0}^{2} (3 x - 1)^{3} d x

52

Solution steps

\int_{0}^{2} (3 x - 1)^{3} d x

Apply u-substitution

: \int_{- 1}^{5} \frac{u ^{3}}{3} d u

= \int_{- 1}^{5} \frac{u ^{3}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{- 1}^{5} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 1}^{5}

= \frac{1}{3} [\frac{u ^{4}}{4}]_{- 1}^{5}

Compute the boundaries:

156

= \frac{1}{3} \cdot 156

Simplify

= 52

\int_{0}^{1} (2 t)^{2} d t

\int_{2}^{5} x (x - 3)^{4} d x

\int_{a}^{b} 4 x^{5} (ln (x^{3})) d x

\int_{- 5}^{- 10} 1 d x

\int_{2}^{4} x^{3} - x^{5} d x

What is the integral from 0 to 2 of (3x-1)^3 ?
The integral from 0 to 2 of (3x-1)^3 is 52